Help with a distance problem?

The problem is easily solved if you determine how far the Smiths traveled before the Garcias started, which can be determined as follows:

d = r(∆t) = (48 miles/hr)(20 minutes)(1 hr/60 minutes) = 16 miles

The differential speed between the Smiths and Garcias is:

∆r = 60 - 48 - 12 miles/hr

The Garcias are effectively traveling 12 miles/hr faster than the Smiths, and it's this differential speed that must cover the 16 miles separating them. The time it takes to cover 16 miles traveling a differential speed of 12 miles/hr is:

t = (16 miles)/(12 miles/hr) = 4/3 hours (1 hour, 20 minutes)

Thus, the time when the Garcias caught up with the Smiths is:

t = 8:20 + 1:20 = 9:40AM

At 8:00 am the smiths left a campground, driving at 48 mi/hr. At 8:20 am the Garcias left the same campground and followed the same route, driving at 60 mi/hr. At what time did they overtake the Smiths?

Find the distance each family traveled from the time that they started.

The equations are based on d = r • t

Ds = 48(t - 8)

Dg = 60(t - 8⅓)

When the families catch up to each other, they are the same distance from their starting point.

Ds = Dg

48(t - 8) = 60(t - 8⅓)

▬▬▬▬▬▬▬▬▬▬▬ Now solve for t, the time at which they meet.

48t - 384 = 60t - 500

116 = 12t

9⅔ = t

So the cars will be at the same point at 9:20 am.